finding the number of solutions $9z^4=\sin^2(z)$ in a complex sector I want to find the number of solution of the equation:
$9z^4 = \sin^2(z)$
in the complex sector: $S=(z\in \Bbb C|-1\le\Im(z)\le 1 )$
I tried to wrote down: $f(z) = \sin^2(z), g(z)=-9z^4$
so we need to find how many zeros are of the function:
$h(z)=f(z)+g(z)$ in the sector $S$.
I want somehow to use Rouche theorem, but this theorem talk only about circles and not sectors of this kind, on the other hand I found out that: $ 0 \le|f(z)|\le 2.381$ but $|g(z)|\le9(\Re(z)^2+1)$
it not helping me show that $|g(z)| > |f(z)|$, and so using Rouche theorem tell us that $g(z)$ and $h(z)$ have the same numbers of zeros (in our case 4).
 A: The main idea is to recognize that with the bound on the imaginary part of $z$ the value of $\sin z$ over the narrow strip can not deviate too much from the sine over the real axis, especially in what concerns the boundedness. And indeed, for any $z=x+iy$ with $|y|=|Im (z)|<1$ we get the bound 
$$
|\sin z|=|\sin x\cosh y+i\cos x\sinh y|= \sqrt{\sin^2 x\cosh^2 y+\cos^2 x\sinh^2 y}\le \cosh y\le \cosh(1)<2
$$
On the other hand, the 4th power grows unbounded with the real part, so that at some point it will dominate the other side. Because of the large factor $9$ it happens rather early.
Indeed, for any $|z|>R=1$ with $Im(z)<R=1$ the function value of $f(z)=9z^4-\sin^2z$ is bounded below by
$$
|f(z)|\ge 9R^4 - \cosh^2(R)=9 - \cosh^2(1)>5>0,
$$
so that there will be no roots of the requested type outside the unit circle.
Now for the roots inside the unit circle consider $g(z)=\sin z^2$. Then $f(z)+g(z)=9z^4$ and $f(z)$ have the same number of roots inside the unit circle by Rouché as for $|z|=r=1$
$$
|f(z)+g(z)|=9r^4=9>\sinh^2(1)=\sinh^2(r)\ge |\sin z|^2=|g(z)|.
$$

One can extend the first argument up to a strip width of $|Im(z)|<R=5$, and also a circle of that radius $R=5$, and the second down to a radius of $r=0.35$ enclosing the double root at $0$ and the other two $\approx\pm\frac13$.
